Question

A projectile is thrown with speed u making angle `theta` with horizontal at `t =0`. It just crosses two points of equal height at time t = 1 s and t = 3 s respectively. Calculate the maximum height attained by it ? (g = `10 m//s^(2)`)

Answer 1

Displacement in y direction `y= u_y xx 1 - (1)/(2)g xx (1)^(2) = u_y xx 3 - (1)/(2) g(3)^(2) rArr u_y = 2g = 20 m//s`
Maximum height attained `h_(max) = (u_y^(2))/(2g) = 20`m.